The equation of a circle $C$ is $x^2+y^2-4x+2y+1 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Solution: To find the equation in standard form, complete the square. $(x^2-4x) + (y^2+2y) = -1$ $(x^2-4x+4) + (y^2+2y+1) = -1 + 4 + 1$ $(x-2)^{2} + (y+1)^{2} = 4 = 2^2$ Thus, $(h, k) = (2, -1)$ and $r = 2$.